You have two decks of cards: a 52 card deck (26 black, 26 red) and a 26 card deck (13 black, 13 red).
- You randomly draw two cards and win if both are the same color. Which deck would you prefer?
- What if the 26 card deck was randomly drawn from the 52 card deck? Which deck would you prefer then?
The first question is straightforward. By symmetry, $$P(\text{same color}) = 2P(\text{two red}) = 2P(\text{second red}|\text{first red})P(\text{first red}).$$ In the 52 card deck, the probability is thus $2 \cdot\frac{25}{51}\frac 12 = \frac{25}{51}$ while in the 26 card deck, it is $2 \cdot \frac{12}{25}\frac 12 = \frac{12}{25}$, and since $\frac{25}{51}> \frac{12}{25}$ the first deck has higher winning odds.
For the second question, here's my approach. Let $R$ be a r.v. modelling the number of red cards in the smaller deck. $R$ follows a hypergeometric distribution: $$P(R=r) = \frac{\binom{26}{r}\binom{26}{26-r}}{\binom{52}{26}},$$ thus $$\begin{align} P(\text{same color}) &= P(\text{two red})+P(\text{two black}) \\ &= \sum_{r=0}^{26} P(\text{second red}|\text{first red, }R=r)P(\text{first red}|R=r)P(R=r) + P(\text{second black}|\text{first black, }R=r)P(\text{first black}|R=r)P(R=r) \\ &=\sum_{r=0}^{26} (\frac{r-1}{25} \frac{r}{26} + \frac{25-r}{25} \frac{r}{26})\frac{\binom{26-r}{r}) \binom{26}{26-r}}{\binom{52}{26}} \\ &= \frac{2}{25\cdot 26} E[R(R-1)] = \frac{2}{25\cdot 26}(V[R]+E[R]^2-E[R]) = \frac{25}{51} \end{align} $$ This is the same probability as for the full deck ! I'm very surprised with this result, I'd like to see an intuitive explanation or a shorter proof that doesn't involve as many computations.
Posting my comment as an answer:
If you first pick $26$ cards out of $52$ and then pick two random cards out of those, it's the same as picking $26$ cards out of $52$ and then taking first two of them because all orderings of $26$ cards are equally likely. Which in turn gives the same result as picking $2$ cards out of $52$ and then not picking $24$ others.